| The a(j) recurrence is a(1)=1; a(2)=11; a(t+2)=12*a(t+1)-a(t)
resulting in terms 1, 11, 131, 1561
The b(j) recurrence is b(1)=1; b(2)=13; b(t+2)=12*b(t+1)-b(t)
resulting in terms 1, 13, 155, 1847
The n(j) recurrence is n(0)=n(1)=0; n(2)=24; n(t+3)=143*(n(t+2)-n(t+1))+n(t)
resulting in terms 0, 0, 24, 3432, 487344 as listed above
G.f.: -24*x^2/((x-1)*(x^2-142*x+1)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 20 2009]
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